All categories

3 years ago

What is the net force on this object?

Physics

2 answers:

Andrew [12]3 years ago

8 0

22 newtons is the net force

zheka24 [161]3 years ago

8 0

8 Newton’s is the net force because normal and gravity cancel each other. The applied is more than friction, 8 N more. So the body moves with a net force of 8 N

You might be interested in

Ow much charge flows from a 12.0 v battery when it is connected across a completely discharged 18.0 μf capacitor

Trava [24]

The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 12.0V (assuming circuit resistance is negligable) and it has a capacitance of 18.0μf or 18.0x10^-6f, therefore charge equals (18.0x10^-6)x12=2.16x10^-4C (Coulombs).

5 0

3 years ago

Three resistors connected in series have potential differences across them labeled /\V1 , /\V2 , and /\V3. What expresses the po

Brrunno [24]

Answer:

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Explanation:

We are given that three resistors R1, R2 and R3 are connected in series.

Let

Potential difference across $R_1=\Delta V_1$

Potential difference across $R_2=\Delta V_2$

Potential difference across $R_3=\Delta V_3$

We know that in series combination

Potential difference , $V=V_1+V_2+V_3$

Using the formula

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Hence, this is required expression for potential difference.

3 0

3 years ago

All the steps of the scientific method rely on valid observations. <br> true or false

butalik [34]

Well, basically because the observations can help you out during the experiment.

4 0

4 years ago

Read 2 more answers

8 cups water a day is that measuring cups?

Gwar [14]

No, it is 8 eight-once cups of water.
Hope it helps! :)

4 0

3 years ago

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte

Lelechka [254]

Answer:

$2.25\mu C$

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

$Q=9.0 \mu C=9.0\cdot 10^{-6}C$ charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

$C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F$

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

$Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C$

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from $9.0 \mu C$ to $11.25 \mu C$ . Therefore, the additional charge that moved to the positive plate is

$\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C$

5 0

3 years ago